Question

Of 2,000 students surveyed, 80% indicated that they commute to school each day. of those that commute to school each day, 83% spend more than $25 a week on gasoline. of those that do not commute to school each day, 95% indicated that they do not spend more than $25 a week on gasoline. a student surveyed is chosen at random. 4. what is the probability that this student does spend more than $25 a week on gas, given that he/she does commute to school each day?

Answer 1

Commute to school 
= 80% of 2000
= 0.8 x 2000
= 1600

Commute to school and spend more than $25 on gasoline
= 83% of 1600
= 0.83 x 1600
= 1328

Given that he/she commute to school,
P(Spend more than $25) = 1382/1600 = 691/800

Answer: 691/800

Answer 2

Using conditional probability, it is found that there is a 0.83 = 83% probability that this student does spend more than $25 a week on gas, given that he/she does commute to school each day.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

• P(B|A) is the probability of event B happening, given that A happened.
• $P(A \cap B)$ is the probability of both A and B happening.
• P(A) is the probability of A happening.

In this problem:

• Event A: Commute to school each day.
• Event B: Spends more than $25 a week.

80% indicated that they commute to school each day. of those that commute to school each day, hence $P(A) = 0.8$.

Of those, 83% spend more than $25 a week on gasoline, hence $P(B|A) = 0.83$.

Thus, 0.83 = 83% probability that this student does spend more than $25 a week on gas, given that he/she does commute to school each day.

A similar problem is given at https://brainly.com/question/14398287